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Chess and math

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  • EXTREMAL PROBLEMS

    These transparencies were used for the second lecture in a series

    of two lectures (the rst given by Noam Elkies) given at M.I.T.

    in January, 2000, and are based on a book in preparation. They

    are available at

    www-math.mit.edu/rstan/trans.html#chess

    Construction tasks.

    Most number of moves by the eightWhite

    ocers (with no Black units on the board)

    M. Bezzel, 1849

    Z Z Z Z

    Z S Z Z

    Z M Z Z

    Z Z Z S

    Z ABM Z

    ZQZ Z Z

    Z J Z Z

    Z Z Z Z

    One hundred moves by eight White pieces.

    Proved by E. Landau in 1899 to be the maximum possi-

    ble.

    1

  • Kling (1849) asked: Can the eight White ocers control

    (guard) every square, occupied or not?

    Can only be done with bishops of the same color (unique

    up to symmetry).

    Kling, 1849

    RZ Z Z Z

    Z Z Z Z

    ZBZ L Z

    Z ZNZ Z

    Z ZNZ Z

    Z J ZBZ

    Z Z Z Z

    Z Z Z ZR

    All squares controlled

    Later proved impossible with bishops on squares of op-

    posite colors. Then the maximum is 63 squares (144 so-

    lutions up to symmetry).

    2

  • Maximum (known) number of checkmates in one move:

    H. Pollmacher and ve others, 1859

    Z ZNZrZ

    O ZPJ ZP

    Z Z Z Z

    ZPZkAQZ

    ZRZ Z Z

    ZBZNZ Z

    Z ZPZ Z

    Z ZRZ Z

    47 mates in one move

    3

  • Maximum number of forced checkmates in one move:

    H. H. Cross, 1936

    Z Z Z Z

    ZbZ Z Z

    pZNZ Z Z

    O s Z Z

    PZ ZKOrZ

    LBj MRO

    PZ S O Z

    Z Z A Z

    29 forced mates in one move

    4

  • The ultimate no-brainer:

    Noam Elkies, 1994

    KAkZ Z Z

    O O Z Zp

    ZPo Z O

    O Z Z Zp

    ZpZ ZrO

    Z O Z oB

    Z Z ZPZ

    Z Z Z Z

    Mate in 7

    Can the number 7 be increased without promoted pieces

    on the board? With promoted pieces a mate in 10 is

    possible (Elkies).

    5

  • Ultimate mutual Zugzwang:

    H. Hunerkopf, 1972

    KAQZ Z Z

    oPO Z Z

    qj o Z Z

    apZPZ Z

    o O Z Z

    ZPZ Z Z

    Z Z Z Z

    Z Z Z Z

    Whoever moves must be mated in one

    Can the number 12 of White and Black initial moves be

    increased?

    Open. For what (m;n) does there exist a legal position

    such that if White moves rst then Blackmust checkmate

    White in exactly m moves, while if Black moves rst

    then White must checkmate Black in exactly n moves.

    (Known only for m = n = 1, and for m = n = 2 using

    promoted pieces (Elkies).)

    6

  • Synthetic games.

    Shortest game ending ending in checkmate (Fool's Mate):

    eight \essentially equivalent" solutions in 2.0 moves (i.e.,

    two White and two Black moves). Two solutions have

    the nal position:

    rmbZkans

    opopZpop

    Z ZpZ Z

    Z Z Z Z

    Z Z ZPl

    Z Z ZPZ

    POPOPZ O

    SNAQJBMR

    Fool's Mate

    7

  • Shortest game ending with checkmate by capture: 2.5

    moves (two essentially equivalent solutions).

    rmbl ans

    opopjpop

    Z Z Z Z

    Z Z L Z

    Z Z Z Z

    Z Z O Z

    POPO OPO

    SNA JBMR

    Mate by capture

    8

  • Construct a legal game of chess in which Black mates

    White by discovered check on Black's fourth move.

    Construct a legal game of chess in which Black makes

    no captures and Black stalemates White on Black's

    twelfth move.

    Construct a legal game of chess in which Black mates

    White on Black's fth move by promoting a pawn to

    a bishop.

    Construct a legal game of chess in which Black mates

    White on Black's fth move by promoting a pawn to

    a knight.

    9

  • Length records.

    Let n be the largest integer for which there exists a dual-

    free mate in n, i.e, in a legal position White is to check-

    mate Black in n move. At least one Black defense forces

    each White move uniquely. All claims to records derive

    from:

    W. Jrgensen, 1976

    Z Z ZnZ

    o ANZ Zb

    pZpZ Zno

    o o Z oq

    Z Z Zks

    J O ZrZp

    ZPOQZ Z

    Z Z Z Z

    Mate in 200

    Longest one with no doubt about soundness: mate in 208

    by A. Cheron (1979). A mate in 226 by Jrgensen himself

    may be sound.

    See Appendix for solution to above problem.

    10

  • What about mate in n allowing duals (i.e., at least one

    Black defense requires n moves by White, but they don't

    have to be uniquely determined)? Without promoted

    force, the record for a long time was n = 257, by O.

    T. Blathy (1889). Recently Ken Thompson reported the

    unique longest mate with RN vs. NN.

    Ken Thompson (& computer), 1999 or 2000

    M Z Z Z

    ZRJ Z Z

    Z Z m Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z m Z

    Z Z Z j

    Mate in 262

    This can be extended to at least mate in 265 with a

    prequel.

    Conclusion: Human beings will never be able to un-

    derstand chess completely.

    11

  • Longest dual-free mate where White has only a queen,

    based on mate in 127 by O. T. Blathy.

    J. Halumbirek (after O. T. Blathy), 1955

    Z Z Z Z

    o Z Z Zp

    Z Z Z o

    o Z Z Zp

    bZ L Z o

    J Z Z Zp

    pZrZ Z Z

    skZ ZnZn

    Mate in 130

    Other dual-free mate records:

    rook minimal: 69 moves

    bishop minimal: 38 moves

    knight minimal: 42 moves

    pawn minimal: 72 moves

    12

  • ENUMERATIVE PROBLEMS

    Let P be a chess position. Let f(n) be the number of dif-

    ferent games (or games ending in checkmate, or \games"

    where only White moves, etc.) starting from P (with

    or without specifying who moves rst). Transfer-matrix

    method )

    Theorem. Let F (x) =

    P

    1

    n=0

    f(n)x

    n

    . Then F (x) is

    a rational function (quotient of two polynomials).

    13

  • Example (Elkies). In the following position, let f(n)

    be the number of ways Black can make n consecutive

    moves, followed by a checkmate in one move by White.

    (Black may not move into check.)

    Z Z ZQZ

    Z Z Z Zp

    Z Z Z j

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z ZK

    Serieshelpmate in n: how many solutions?

    Then

    1

    X

    n=0

    f(n)x

    n

    =

    x

    5

    (2 + 5x 4x

    2

    2x

    3

    )

    (1 x

    2

    )(1 2x

    2

    )(1 3x

    2

    + x

    4

    )

    = 2x

    5

    + 5x

    6

    + 8x

    7

    + 28x

    8

    + 24x

    9

    + 108x

    10

    + 66x

    11

    +

    f(2m) = 3 2

    m+2

    + F

    2m+3

    ; m > 0

    f(2m + 1) = 2(F

    2m1

    1);

    where F

    n

    is a Fibonacci number:

    F

    1

    = F

    2

    = 1; F

    n+1

    = F

    n

    + F

    n1

    :

    14

  • Let P be the initial position. Let f(n) be the number of

    games in n single moves, and g(n) the number ending in

    checkmate. Then

    (f(0); f(1); : : : ; f(8)) = (1; 20; 400; 8902; 197281; 4865609;

    119060324; 3195901860;

    84998978956; 2439530234167):

    (g(4); g(5); g(6)) = (8; 347; 10828);

    the latter computed in 1897.

    15

  • Classical ballot problem. Let C

    n

    be the number of

    ways 2n voters can vote sequentially for two candidates

    A and B, so that each receives n votes and A never trails

    B during the voting. E.g., C

    3

    = 5:

    AAABBB AABABB AABBAA ABAABB ABABAB:

    C

    n

    is the Catalan number

    C

    n

    =

    1

    n + 1

    2n

    n

    :

    For 66 combinatorial interpretations, see Exercise 6.19 of

    R. Stanley, Enumerative Combinatorics, vol. 2.

    16

  • K. Vaisanen, 1992

    Z Z ZkZ

    o Z Z Z

    pZ ZPO Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z ZK

    A Z Z Z

    Z Z Z Z

    Serieshelpmate in 19: how many solutions?

    Serieshelpmate in n: Black moves rst and makes n

    consecutive move. White then checkmates Black in one

    move. Black and White are cooperating. Black may not

    move into check or check White (except possibly on his

    last move).

    17

  • ss

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    s

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @@

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @

    @@

    P

    1

    a5

    P

    1

    a4

    P

    2

    a5

    P

    1

    a3

    P

    1

    a2

    P

    1

    a1N

    N

    1

    b3

    N

    1

    c5

    N

    1

    d7

    N

    1

    f8

    N

    1

    h7

    P

    2

    a4

    P

    2

    a3

    P

    2

    a2

    P

    2

    a1N

    N

    2

    b3

    N

    2

    c5

    N

    2

    d7

    N

    2

    f8

    Figure 1: Solution poset

    Number of solutions = C

    9

    =

    1

    10

    18

    9

    = 4862.

    18

  • RETROGRADE ANALYSIS

    R. Smullyan, 1979

    Z ZkZ Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z O O Z

    Z Z J Z

    Monochrome chess. A White bishop stands at e3 or e4. Which?

    Monochrome chess: every move is to a square of the

    same color (so e.g. knights never move).

    19

  • N. Hoeg, 1916

    Z Z Z Z

    Z ZKZ Z

    Z ZPj Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    A Z Z Z

    What was the last move?

    20

  • N. Petrovic, 1954

    Z Z ZKZ

    Z Z Z ZB

    Z ZPj Z

    Z Z Z Z

    Z Z ZQZ

    Z Z L Z

    Z Z Z Z

    A Z Z Z

    The last six single moves?

    21

  • J. Furman, 1973

    ZBZ Z Z

    orZpOpo

    kZ Z ZpZ

    O Z LpZ

    Ko Z Z Z

    ZrO Z ZP

    POpZPZ O

    S Z M ZR

    White's Rooks have exchanged places

    22

  • G. Schweig, 1938

    rZblkans

    opo Zpop

    Z o Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    POPOPOPO

    S AQJBMR

    Position after the 4th move of Black. How did the game go?

    I.e., proof game in 4.0 moves

    23

  • Phoenix theme:

    R. Muller, 1985

    mblkans

    o o opop

    Z o Z Z

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    OPO OPO

    SNAQJBMR

    Proof game in 6.5 moves

    24

  • Tibor Orban, 1976

    rmblkans

    opZ Zpop

    ZpZpZ Z

    Z Z Z Z

    Z ZPZ Z

    Z Z Z Z

    POPO OPO

    SNAQJ MR

    Proof game in exactly 4.0 moves

    Solution in 3.5 moves is easy!

    25

  • G. Donati, 1998

    rm Z Z Z

    opZpopo

    oqj m o

    Z Z Z Z

    Z Z Z Z

    Z Z Z Z

    PO OPOPO

    SNAQJBMR

    Proof game in 13.5 moves

    A nice puzzle!

    26

  • APPENDIX

    Solution to mate in 200 by Jrgensen: We sketch the solution to this complex

    problem following C. J. Morse. Ample study is necessary for full apprecia-

    tion. Certain maneuvers are used repeatedly and are indicated after their

    rst appearance by ellipses (: : :). The solution begins 1. Qe6+ Rf5 2. Bh2.

    Suppose that White's king were now at d1 and that the ve Black queenside

    pawns had no moves. If Black plays Kf3, then Qe2 is mate. The only alter-

    native for Black is to move one of his knights. If he moves the knight on g6,

    then Ne5 is mate. If Black instead moves the knight on g8, then White mates

    in two by Nf6+ Kf3, Qe2 mate. Hence White's goal is to manuever his king

    to d1 while capturing or immobilizing Black's queenside pawns. However,

    White may not move his king to the fourth rank, since then Black can move

    his king o the fourth rank with discovered check. The solution continues

    2: : : Kf3 (other moves shorten the solution) 3. Qe3+ (not 3. Qxf5+ Kg2!,

    and Black can free his queen. It is essential to keep Black's pieces bottled

    up.) Kg2 (of course not Kg4 4. Qg3 mate) 4. Qg1+ Kf3 5. Qf1+ Kg4 (not

    Ke4 6. Qd3 mate) 6. Qe2+ Rf3 7. Qe6+ Rf5 8. Ka2 (not 8. Ka4 Kf3+! nor

    8. c4 Kf3 leading to 12. Qe2+ Rf3+!) Kf3 : : : 14. Kc1 Kf3 : : : 20. Kd1 a4

    (not Kf3 21. Ke2 mate, nor c4 which allows 21. Ke1 a4 22. Kd1) 21. Kc1

    (not 21. Ke1 a3!, and Black will be able to queen his pawn and check White

    before White can play Ke4 followed by either Qxf5 mate or by mating with

    the knight if Black moves a knight) Kf3 : : : 27. Kd1 a5 (c4 is playable here

    and at similar positions up to move 92, but not a3 which allows 65. Ke1) 28.

    Kc1 Kf3 : : : 34. Kd1 a6 (here a3 is an alternative as well as c4) 35. Kc1 Kf3

    : : : 41. Kd1 a3 42. Kc1 Kf3 : : : 48. Kb1 (not 48. Kd1 a2!) Kf3 : : : 54. Ka2 Kf3

    : : : 60. Kxa3 Kf3 : : : 66. Kb2 Kf3 : : : 72. Kc1 Kf3 : : : 78. Kd1 a4 : : : 85. Kd1

    a5 : : : 92. Kd1 c4 93. Kc1 Kf3 : : : 96. Qf1+ Kg4 (not Ke4 97. e3+ Kd5 98.

    Qxf5+ leading to short mate) : : : 99. Kd1 c5 : : : 106. Kd1 a3 107. Kc1 Kf3

    : : : 113. Kb1 Kf3 : : : 116. Qf1+ Ke4 (this can be played without shortening

    the solution when the White king is not guarding d2) 117. Qxc4+ (not 117.

    d3+ Ke3 118. Bg1+ Kd2!) Kf3 118. Qf1+ Kg4 : : : 120. Qe6+ Rf5 121 Ka2

    Kf3 : : : 127. Kxa3 Kf3 : : : 145. Kd1 a4 : : : 152. Kd1 c4 : : : 159. Kd1 (not

    159. Kb2, leading to 168. Qf1+ Ke4 and 173. Kxa4 Kf3+!) a3 160. Kc1 Kf3

    : : : 166. Kb1 Kf3 : : : 169. Qf1+ Ke4 (this can be delayed until move 187) : : :

    174. Ka2 Kf3 : : : 180. Kxa3 Kf3 : : : 198. Kd1 (nally reaching the desired

    position!) Ng8 moves 199. Nf6+ Kf3 200. Qe2 mate!

    27